Posts Tagged ‘ideal gas’

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

The diffusion constant relation to the probability current is referred to as Fick’s law

with which we can cast the probability diffusion identity into a continuity equation form

In 3D (with the Maxwell distribution frictional term), this takes the form

Maxwell distribution

Add a frictional term to the velocity space diffusion current

For steady state the continity equation leads to

We also find

and identify

Hamilton’s equations

SHO

Quantum energy eigenvalues

Liouville’s theorem

Regardless of whether we have a steady state system, if we sit on a region of phase space volume, the probability density in that neighbourhood will be constant.

Ergodic

A system for which all accessible phase space is swept out by the trajectories. This and Liouville’s threorm allows us to assume that we can treat any given small phase space volume as if it is equally probable to the same time evolved phase space region, and switch to ensemble averaging instead of time averaging.

Disclaimer

Fermi gas

Review

Continuing a discussion of [1] section 8.1 content.

We found

With no spin

Fig 1.1: Occupancy at low temperature limit

Fig 1.2: Volume integral over momentum up to Fermi energy limit

gives

This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intended. We see that both achieve the same result}, where

Moving on

with

this gives

Over all dimensions

so that

Again

Example: Spin considerations

{example:basicStatMechLecture16:1}{

This gives us

and again

}

High Temperatures

Now we want to look at the at higher temperature range, where the occupancy may look like fig. 1.3

Fig 1.3: Occupancy at higher temperatures

so that for large we have

Mathematica (or integration by parts) tells us that

so we have

Introducing for the thermal de Broglie wavelength,

we have

Does it make any sense to have density as a function of temperature? An inappropriately extended to low temperatures plot of the density is found in fig. 1.4 for a few arbitrarily chosen numerical values of the chemical potential , where we see that it drops to zero with temperature. I suppose that makes sense if we are not holding volume constant.

Fig 1.4: Density as a function of temperature

We can write

or (taking (and/or volume?) as a constant) we have for large temperatures

The chemical potential is plotted in fig. 1.5, whereas this function is plotted in fig. 1.6. The contributions to from the term are dropped for the high temperature approximation.

Disclaimer

Question: Sackur-Tetrode entropy of an Ideal Gas

Find the temperature of this gas via . Find the energy per particle at which the entropy becomes negative. Is there any meaning to this temperature?

Answer

Taking derivatives we find

or

The energies for which the entropy is negative are given by

or

In terms of the temperature this negative entropy condition is given by

or

There will be a particle density for which this distance will start approaching the distance between atoms. This distance constrains the validity of the ideal gas law entropy equation. Putting this quantity back into the entropy eq. 1.1.1 we have

We see that a positive entropy requirement puts a bound on this distance (as a function of temperature) since we must also have

for the gas to be in the classical domain. I’d actually expect a gas to liquefy before this transition point, making such a low temperature nonphysical. To get a feel for whether this is likely the case, we should expect that the logarithm argument to be

at the point where gasses liquefy (at which point we assume the ideal gas law is no longer accurate) to be well above unity. Checking this for 1 liter of a gas with atoms for hydrogen, helium, and neon respectively we find the values for eq. 1.1.10 are

At least for these first few cases we see that the ideal gas law has lost its meaning well before the temperatures below which the entropy would become negative.

Question: Ideal gas thermodynamics

An ideal gas starts at in the pressure-volume diagram (x-axis = , y-axis = ), then moves at constant pressure to a larger volume , then moves to a larger pressure at constant volume to , and finally returns to , thus undergoing a cyclic process (forming a triangle in plane). For each step, find the work done on the gas, the change in energy content, and heat added to the gas. Find the total work/energy/heat change over the entire cycle.

Answer

Our process is illustrated in fig. 1.1.

Fig 1.1: Cyclic pressure volume process

Step 1
This problem is somewhat underspecified. From the ideal gas law, regardless of how the gas got from the initial to the final states, we have

So a volume increase with fixed implies that there is a corresponding increase in . We could have for example, an increase in the number of particles, as in the evaporation process illustrated of fig. 1.2, where a piston held down by (fixed) atmospheric pressure is pushed up as the additional gas boils off.

Fig 1.2: Evaporation process under (fixed) atmospheric pressure

Alternately, we could have a system such as that of fig. 1.3, with a fixed amount of gas is in contact with a heat source that supplies the energy required to induce the required increase in temperature.

Fig 1.3: Gas of fixed mass absorbing heat

Regardless of the source of the energy that accounts for the increase in volume the work done on the gas (a negation of the positive work the gas is performing on the system, perhaps a piston as in the picture) is

Let’s now assume that we have the second sort of configuration above, where the total amount of gas is held fixed. From the ideal gas relations of eq. 1.0.12.12, and with , , and , we have

The change in energy of the ideal gas, assuming three degrees of freedom, is

The energy balance then requires that the total heat absorbed by the gas must include that portion that has done work on the system, plus the excess kinetic energy of the gas. That is

Step 2

For this leg of the cycle we have no work done on the gas

We do, however have a change in energy. The energy of the gas is

With , the change of energy of the gas, the total heat absorbed by the gas, is

Step 3

For the final leg of the cycle, the work done on the gas is

This is positive this time
Unlike the first part of the cycle, the work done on the gas is positive this time (work is being done on the gas to both compress it). The change in energy of the gas, however, is negative, with the difference between final and initial energy being

The simultaneous compression and the pressure reduction require energy to be removed from the gas. We must have a negative change in heat , with heat emitted in this phase of the cycle. This can be verified explicitly

Changes over the complete cycle.

Summarizing the results from each of the phases, we have

Summing the changes in the work we have

This is the area of the triangle, as expected. Since it is positive, there is net work done on the gas.

We expect the energy changes to sum to zero, and this can be verified explicitly finding

With net work done on the gas and no change in energy, there should be no net heat absorption by the gas, with a total change in heat that should equal, in amplitude, the total work done on the gas. This is confirmed by summation

Question: Adiabatic process for an Ideal Gas

Show that when an ideal monoatomic gas expands adiabatically, the temperature and pressure are related by

Answer

From (3.34b) of [1], we find that the Adiabatic condition can be expressed algebraically as

With

this is

Dividing through by , this becomes a perfect differential, and we can integrate

Question: Rotation of diatomic molecules ([2] problem 3.6)

In our first look at the ideal gas we considered only the translational energy of the particles. But molecules can rotate, with kinetic energy. The rotation motion is quantized; and the energy levels of a diatomic molecule are of the form

where is any positive integer including zero: . The multiplicity of each rotation level is .

a

Find the partition function for the rotational states of one molecule. Remember that is a sum over all states, not over all levels — this makes a difference.

b

Evaluate approximately for , by converting the sum to an integral.

c

Do the same for , by truncating the sum after the second term.

d

Give expressions for the energy and the heat capacity , as functions of , in both limits. Observe that the rotational contribution to the heat capacity of a diatomic molecule approaches 1 (or, in conventional units, ) when .

e

Sketch the behavior of and , showing the limiting behaviors for and .

Answer

a. Partition function

To understand the reference to multiplicity recall (section 4.13 [1]) that the rotational Hamiltonian was of the form

where the eigenvectors satisfied

\begin{subequations}

\end{subequations}

and , where is a positive integer. We see that is of the form

and our partition function is

We have no dependence on in the sum, and just have to sum terms like fig 1, and are able to sum over trivially, which is where the multiplicity comes from.

Fig 1: Summation over m

To get a feel for how many terms are significant in these sums, we refer to the plot of fig 2. We plot the partition function itself in, truncation at terms in fig 3.

Fig 2: Plotting the partition function summand

Fig 3: Z_R(tau) truncated after 30 terms in log plot

b. Evaluate partition function for large temperatures

If , so that , all our exponentials are close to unity. Employing an integral approximation of the partition function, we can somewhat miraculously integrate this directly

c. Evaluate partition function for small temperatures

When , so that , all our exponentials are increasingly close to zero as increases. Dropping all the second and higher order terms we have

d. Energy and heat capacity

In the large domain (small temperatures) we have

The specific heat in this domain is

For the small (large temperatures) case we have

The heat capacity in this large temperature region is

which is unity as described in the problem.

e. Sketch

The energy and heat capacities are roughly sketched in fig 4.

Fig 4: Energy and heat capacity

It’s somewhat odd seeming that we have a zero point energy at zero temperature. Plotting the energy (truncating the sums to 30 terms) in fig 5, we don’t see such a zero point energy.

Fig 5: Exact plot of the energy for a range of temperatures (30 terms of the sums retained)

That plotted energy is as follows, computed without first dropping any terms of the partition function

To avoid the zero point energy, we have to use this and not the truncated partition function to do the integral approximation. Doing that calculation (which isn’t as convenient, so I cheated and used Mathematica). We obtain

This approximation, which has taken the sums to infinity, is plotted in fig 6.

Fig 6: Low temperature approximation of the energy

From eq. 1.0.12, we can take one more derivative to calculate the exact specific heat

Question: Classical gas partition function

Show that this leads to the expected result for the thermal average energy.

Answer

Let’s use the adjustment technique from the text for the partition case and write

with as above. This gives us

Question: Two state system

[1] problem 3.1.

Find an expression for the free energy as a function of of a system with two states, one at energy and one at energy . From the free energy, find expressions for the energy and entropy of the system.

Answer

Our partition function is

The free energy is just

The entropy follows immediately

The energy is

This is

These are all plotted in (Fig 1).

Fig1: Plots for two state system

\imageFigure{kittelCh3Problem1PlotsFig1}{Plots for two state system}{fig:kittelCh3Problem1Plots:kittelCh3Problem1PlotsFig1}{0.2}

Question: Magnetic susceptibility

[1] problem 3.2.

Use the partition function to find an exact expression for the magnetization and the susceptibility as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization, found by other means, was , where is the particle concentration. Find the free energy and express the result as a function only of and the parameter . Show that the susceptibility is in the limit .

Answer

Our partition function for a unit volume containing spins is

so that the Free energy is

The energy, magnetization and magnetic field were interrelated by

This gives us

so that

For , the cosh term goes to unity, so we have

as desired.

With , or , the free energy is

That last expression isn’t particularly illuminating. What was the point of that substitution?

Question: Free energy of a harmonic oscillator

[1] problem 3.3.

A one dimensional harmonic oscillator has an infinite series of equally spaced energy states, with , where is a positive integer or zero, and is the classical frequency of the oscillator. We have chosen the zero of energy at the state . Show that for a harmonic oscillator the free energy is

Note that at high temperatures such that we may expand the argument of the logarithm to obtain . From 1.0.16 show that the entropy is

Answer

I found it curious that this problem dropped the factor of from the energy. Including it we have

So that the partition function is

The free energy is

We see that the contribution of the in the energy of each state just adds a constant factor to the free energy. This will drop out when we compute the entropy. Dropping that factor now that we know why it doesn’t contribute, we can complete the summation, so have, by inspection

Taking derivatives for the entropy we have

Question: Energy fluctuation

[1] problem 3.4.

Consider a system of fixed volume in thermal contact with a reservoir. Show that the mean square fluctuation in the energy of the system is

Here is the conventional symbol for . Hint: Use the partition function to relate to the mean square fluctuation. Also, multiply out the term .

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Liouville’s theorem

We’ve looked at the continuity equation of phase space density

which with

led us to Liouville’s theorem

We define Ergodic, meaning that with time, as you wait for , all available phase space will be covered. Not all systems are necessarily ergodic, but the hope is that all sufficiently complicated systems will be so.

We hope that

In particular for , we see that our continuity equation 1.2.1 results in 1.2.2.

For example in a SHO system with a cyclic phase space, as in (Fig 1).

Fig 1: Phase space volume trajectory

or equivalently with an ensemble average, imagining that we are averaging over a number of different systems

If we say that

so that

then what is this constant. We fix this by the constraint

So, is the allowed “volume” of phase space, the number of states that the system can take that is consistent with conservation of energy.

What’s the probability for a given configuration. We’ll have to enumerate all the possible configurations. For a coin toss example, we can also ask how many configurations exist where the sum of “coin tosses” are fixed.

A worked example: Ideal gas calculation of

gas atoms at phase space points

constrained to volume

Energy fixed at .

With defined implicitly by

so that with Heavyside theta as in (Fig 2).

Fig 2: Heavyside theta

we have

In three dimensions , the dimension of momentum part of the phase space is 3. In general the dimension of the space is . Here

is the volume of a “sphere” in – dimensions, which we found in the problem set to be

Since we have

the radius is

This gives

and

This result is almost correct, and we have to correct in 2 ways. We have to fix the counting since we need an assumption that all the particles are indistinguishable.

Indistinguishability. We must divide by .

is not dimensionless. We need to divide by , where is Plank’s constant.

In the real world we have to consider this as a quantum mechanical system. Imagine a two dimensional phase space. The allowed points are illustrated in (Fig 3).

Fig 3: Phase space volume adjustment for the uncertainty principle

Since , the question of how many boxes there are, we calculate the total volume, and then divide by the volume of each box. This sort of handwaving wouldn’t be required if we did a proper quantum mechanical treatment.